Question

我在我的博客上添加了评论系统，问题是我无法在数据库上插入评论，我猜是因为它无法获取当前博客的帖子ID。从comments.php，数据将传递给post_comments.php。我怎样才能在ajax部分（脚本标签下）传递post_id？这是我第一次使用ajax，所以我对它并不是很好。任何帮助将不胜感激

comments.php：

<form method='post' action="" onsubmit="return post();">
<textarea id="comment" placeholder="Write Your Comment Here....."></textarea>
<br>
<input type="text" id="username" placeholder="Your Name">
<br>
<input type="submit" value="Post Comment">
</form>

<div id="all_comments">
<?php
    $host="localhost";
    $username="root";
    $password="";
    $databasename="comments";

    $connect=mysql_connect($host,$username,$password);
    $db=mysql_select_db($databasename);

    $comm = mysql_query("select name,comment,post_time from comments order by post_time desc");
    while($row=mysql_fetch_array($comm))
    {
        $name=$row['name'];
        $comment=$row['comment'];
        $time=$row['post_time'];
        ?>
        <div class="comment_div"> 
        <p class="name">Posted By:<?php echo $name;?></p>
        <p class="comment"><?php echo $comment;?></p> 
        <p class="time"><?php echo $time;?></p>
        </div>
        <?php  
    }   
    ?>
</div>

post_comments.php

<?php
  $host="localhost";
  $username="root";
  $password="";
  $databasename="comments";

  $connect=mysql_connect($host,$username,$password);
  $db=mysql_select_db($databasename);

  if(isset($_POST['user_comm']) && isset($_POST['user_name']))
   {
        $comment=$_POST['user_comm'];
        $name=$_POST['user_name'];
        $insert=mysql_query("insert into comments values('','$name','$comment',CURRENT_TIMESTAMP)");
        $select=mysql_query("select name,comment,post_time from comments where name='$name' and comment='$comment' ");

    if($row=mysql_fetch_array($select))
    {
        $name=$row['name'];
        $comment=$row['comment'];
        $time=$row['post_time'];
       ?>

    <div class="comment_div"> 
      <p class="name">Posted By:<?php echo $name;?></p>
      <p class="comment"><?php echo $comment;?></p> 
      <p class="time"><?php echo $time;?></p>
    </div>
<?php 
    } 
    exit; 
  }  
?>

Answer 1

您需要传递post_id。因此，您需要将post_id存储到隐藏字段中。并在Ajax电话代码中进行检查。

<强>的comments.php

......
<input type="hidden" id="post_id" name="post_id" value="<?php echo $post_id?>">
<input type="submit" value="Post Comment">
......

<强> post_comments.php

......
if(isset($_POST['user_comm']) && isset($_POST['user_name']) && isset($_POST['post_id']))
   {
        $post_id=$_POST['post_id'];
        // do what ever you wanted to do with post_id 
        $comment=$_POST['user_comm'];

......

如何使用AJAX在评论系统中获取当前页面ID

1 个答案: